Solve (frac{9a^{2}+6ab+b^{2}}{9a^{2}-b^2}-2a^2/3a^2-ab):a^2+2ab+b/b-3a

$$( \frac { 9 a ^ { 2 } + 6 a b + b ^ { 2 } } { 9 a ^ { 2 } - b ^ { 2 } } - \frac { 2 a ^ { 2 } } { 3 a ^ { 2 } - a b } ) : \frac { a ^ { 2 } + 2 a b + b } { b - 3 a }$$

Evaluate

$$-\frac{a+b}{a^{2}+2ab+b}$$

Short Solution Steps

Factor the expressions that are not already factored in \frac{9a^{2}+6ab+b^{2}}{9a^{2}-b^{2}}.

$$\frac{\frac{\left(3a+b\right)^{2}}{\left(3a+b\right)\left(3a-b\right)}-\frac{2a^{2}}{3a^{2}-ab}}{\frac{a^{2}+2ab+b}{b-3a}}$$

Cancel out 3a+b in both numerator and denominator.

$$\frac{\frac{3a+b}{3a-b}-\frac{2a^{2}}{3a^{2}-ab}}{\frac{a^{2}+2ab+b}{b-3a}}$$

Factor the expressions that are not already factored in \frac{2a^{2}}{3a^{2}-ab}.

$$\frac{\frac{3a+b}{3a-b}-\frac{2a^{2}}{a\left(3a-b\right)}}{\frac{a^{2}+2ab+b}{b-3a}}$$

Cancel out a in both numerator and denominator.

$$\frac{\frac{3a+b}{3a-b}-\frac{2a}{3a-b}}{\frac{a^{2}+2ab+b}{b-3a}}$$

Since \frac{3a+b}{3a-b} and \frac{2a}{3a-b} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{3a+b-2a}{3a-b}}{\frac{a^{2}+2ab+b}{b-3a}}$$

Combine like terms in 3a+b-2a.

$$\frac{\frac{a+b}{3a-b}}{\frac{a^{2}+2ab+b}{b-3a}}$$

Divide \frac{a+b}{3a-b} by \frac{a^{2}+2ab+b}{b-3a} by multiplying \frac{a+b}{3a-b} by the reciprocal of \frac{a^{2}+2ab+b}{b-3a}.

$$\frac{\left(a+b\right)\left(b-3a\right)}{\left(3a-b\right)\left(a^{2}+2ab+b\right)}$$

Extract the negative sign in b-3a.

$$\frac{-\left(a+b\right)\left(3a-b\right)}{\left(3a-b\right)\left(a^{2}+2ab+b\right)}$$

Cancel out 3a-b in both numerator and denominator.

$$\frac{-\left(a+b\right)}{a^{2}+2ab+b}$$

To find the opposite of a+b, find the opposite of each term.

$$\frac{-a-b}{a^{2}+2ab+b}$$

Expand

$$-\frac{a+b}{a^{2}+2ab+b}$$

Short Solution Steps

Factor the expressions that are not already factored in \frac{9a^{2}+6ab+b^{2}}{9a^{2}-b^{2}}.

$$\frac{\frac{\left(3a+b\right)^{2}}{\left(3a+b\right)\left(3a-b\right)}-\frac{2a^{2}}{3a^{2}-ab}}{\frac{a^{2}+2ab+b}{b-3a}}$$

Cancel out 3a+b in both numerator and denominator.

$$\frac{\frac{3a+b}{3a-b}-\frac{2a^{2}}{3a^{2}-ab}}{\frac{a^{2}+2ab+b}{b-3a}}$$

Factor the expressions that are not already factored in \frac{2a^{2}}{3a^{2}-ab}.

$$\frac{\frac{3a+b}{3a-b}-\frac{2a^{2}}{a\left(3a-b\right)}}{\frac{a^{2}+2ab+b}{b-3a}}$$

Cancel out a in both numerator and denominator.

$$\frac{\frac{3a+b}{3a-b}-\frac{2a}{3a-b}}{\frac{a^{2}+2ab+b}{b-3a}}$$

Since \frac{3a+b}{3a-b} and \frac{2a}{3a-b} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{3a+b-2a}{3a-b}}{\frac{a^{2}+2ab+b}{b-3a}}$$

Combine like terms in 3a+b-2a.

$$\frac{\frac{a+b}{3a-b}}{\frac{a^{2}+2ab+b}{b-3a}}$$

Divide \frac{a+b}{3a-b} by \frac{a^{2}+2ab+b}{b-3a} by multiplying \frac{a+b}{3a-b} by the reciprocal of \frac{a^{2}+2ab+b}{b-3a}.

$$\frac{\left(a+b\right)\left(b-3a\right)}{\left(3a-b\right)\left(a^{2}+2ab+b\right)}$$

Extract the negative sign in b-3a.

$$\frac{-\left(a+b\right)\left(3a-b\right)}{\left(3a-b\right)\left(a^{2}+2ab+b\right)}$$

Cancel out 3a-b in both numerator and denominator.

$$\frac{-\left(a+b\right)}{a^{2}+2ab+b}$$

To find the opposite of a+b, find the opposite of each term.

$$\frac{-a-b}{a^{2}+2ab+b}$$